Question

The graph of the continuous function g, the derivative of the function f, is shown above. The function g is piecewise linear for -5 ≤ x < 3 and g(x) = 2(x-4)^2 for 3 ≤ x ≤ 6.
(a) If f(1) = 3, what is the value of f(-5) ?

(b) Evaluate $\int_{0}^{6} g(x)\; dx$ .

(c) For -5 < x < 6, on what open intervals, if any, is the graph of f both increasing and concave up? Give a reason for your answer.

(d) Find the x-coordinate of each point of inflection of the graph of f. Give a reason for your answer.

(This was on the no-calculator section of the recently-released AP Calculus AB 2018 exam so I appreciate it if you tried to limit calculator usage)

Answer 1

A) $g=f'$ is continuous, so $f$ is also continuous. This means if we were to integrate $g$, the same constant of integration would apply across its entire domain. Over $0$, we have $g(x)=2x$. This means that


$f_{0$


For $f$ to be continuous, we need the limit as $x\to1^-$ to match $f(1)=3$. This means we must have


$\displaystyle\lim_{x\to1}x^2+C=1+C=3\implies C=2$


Now, over $x$, we have $g(x)=-3$, so $f_{x$, which means $f(-5)=17$.


b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,


$\displaystyle\int_1^6g(x)=4+\int_3^62(x-4)^2\,\mathrm dx=4+6=10$


c) $f$ is increasing when $f'=g>0$, and concave upward when $f''=g'>0$, i.e. when $g$ is also increasing.

We have $g>0$ over the intervals $0$ and $x>4$. We can additionally see that $g'>0$ only on $0$ and $x>4$.


d) Inflection points occur when $f''=g'=0$, and at such a point, to either side the sign of the second derivative $f''=g'$ changes. We see this happening at $x=4$, for which $g'=0$, and to the left of $x=4$ we have $g$ decreasing, then increasing along the other side.


We also have $g'=0$ along the interval $-1$, but even if we were to allow an entire interval as a "site of inflection", we can see that $g'>0$ to either side, so concavity would not change.